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I have this circuit.

enter image description here

I am trying to calculate the voltage between A and E by using Kirchhoff current law.

This is what I did.

First I wrote the equation for the left side, starting at A clockwise:

\$ -V_{B2} - I_{1}R_{3} + I_{2}R_{3} - I_{1}R_{5} + V_{1} - I_{1}R_{1} = 0 \$

plugging in the value and simplifying, I get:

\$ 17I_{1} - 10I_{2} = 8 \$

Now lets talk about I2. I assumed that this current was clockwise, so

\$ I_{2} = -2 A \$

right?

Plugging in this value on the first equation I get

\$ I_{1} = -0.705 A \$

Now I draw the 3 currents for the branches...

enter image description here

\$ i_1 \$ will be equal to \$ I_{1} = -0.705 A \$ \$ i_2 \$ will be equal to \$ I_{2} = -2 A \$

and

\$ i_3 = i_1 - i_2 = 1.294 A \$

so, \$ V_{AE} = 2 + 1.294 \times 10 = 14.94 V \$

I arrived at this value but B2 is killing me. I am not sure if I should add or subtract B2 to get \$ V_{AE} \$

is that correct?

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Adding it as you did is correct. B2 has a voltage drop ( in the A -> E) direction. Equivalently, Ve + 10*1.294 + 2 = Va. Your numbers also check out in Spice

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  • \$\begingroup\$ Thanks for the check. My brain was melting for a second. \$\endgroup\$ – SpaceDog Aug 6 '18 at 2:51

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